Consider the reaction:
2 SO2( g) + O2( g)¡2 SO3( g)
a. If 285.5 m...

Question

Consider the reaction:
2 SO2( g) + O2( g)¡2 SO3( g)
a. If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both
measured at 315 K and 50.0 mmHg), what is the limiting
reactant and the theoretical yield of SO3?

Accepted Answer

Hey everyone, Let's check out this problem at STP, calculate the mass of phosphene that can be formed from the reaction of 42.5 g of phosphorus and 96 liters of hydrogen gas. Now, STP stands for standard temperature and pressure in our standard temperature is equal to 273.15 Kelvin and our standard pressure Is equal to 1 80 M. So the first thing we can do here is balance our equation here. It's not balanced on the left side. We have four phosphorus, so we'll add a four here. And that changes our hydrogen amount on the right side to 12 hydrogen. So we'll add a six here. Now we're going to calculate the amount of product produced from each of the react mints and we'll get to numbers and the lesser amount of product formed is going to be our limiting reactant. And that's going to be the answer here. So let's start off with one of the reactant that we know And that is 42.5 g phosphorus. So we want to go from grams of phosphorus two g of product and we can do that using our balanced equation and our molecular weights. So in one mole Of phosphorus, using our periodic table, we see that phosphorus weighs 31 g and we have four. So that gives us a molecular weight of 124 g and our units will cancel here and we can go from moles of phosphorus two moles of phosphorus in using our multiple ratio in our equation. So in one mole of phosphorus we have four moles of phosphorus Our units cancel. And we want to go from most of phosphorus 2g of phosphorus and we'll do that using the molecular weight of phosphorus. So in one mole of phosphene we have Phosphene is and we have hydrogen and We have three of them in each weighs one. So we get a total of 34 g. Alright our units cancel and we can go ahead and do our math here and when we do that, We get 46 . g of phosphorus. So that's for one of our reactant. We need to do the same thing but we'll do it for our hydrogen gas and for hydrogen gas they tell us we have 96 liters. And so we can use our ideal gas equation to solve for this one. So are ideal gas equation is PV equals N. R. T. And we want to solve for N. Which will give us moles of hydrogen gas and from moles of hydrogen gas we can go to grams of phosphorus in. So let's go ahead and start off by dividing both sides and isolating our N. So we get and is equal to PV over R. T. And we can go ahead and plug in what we know, We know that our pressure is 180 M. Our volume is given 96 leaders, R r is r gas constant 0.08206 leaders atmosphere over more Calvin. And our temperature Is our standard temperature of 273.15 Kelvin. So let's make sure our units cancel. Are ATMs cancel, our leaders cancel and our kelvin's cancel. And we're left with moles. Let me make this a little smaller. Okay, so once we do this we get n is equal to 4. Moles of H two. Now we need to go from moles of H22g of our product phosphene. So we'll do the same thing using dimensional analysis in our balanced equation. So Um are multiple ratio between our hydrogen gas and phosgene is 4-6 in six moles of hydrogen gas. Get this from our balanced equation. We have four moles of phosphene And in one mole of phosgene our molecular weight Is 34 g. Okay, so make sure our units cancel moles of hydrogen gas cancel and moles of phosphate in cancel. And we're left with a weight of 97 grams of phosphorus. So now we can see how much each of our reactant sites produce. And the one that is the lesser amount is called are limiting reactant. So here we have 46.6 g of phosphene is produced from our phosphorus and this is going to be our correct answer. And that is the end of this problem. I hope this was helpful

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